Strength Of A Perfect Crystal
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Strength of a perfect crystal
• Slip is assumed to occur by translation of one plane of atoms over another • Reason able estimate of shear stress required can be done by such a movement in perfect crystal • Therefore consider two planes of atoms subjected to a homogeneous shear stress
• Shear stress is assumed to act in a slip plane along the slip direction • The distance between the atoms in the slip direction is “b” • The spacing between adjacent lattice is “a” • The shear stress cause a displacement “x” in the slip direction between the pair of adjacent lattice planes
• The shear stress is zero when • Two planes are in coincidence, initial stage • When the two planes have move one identity distance • Atoms of the top plane are midway between those of the bottom plane
• Shearing stress is a periodic function of the displacement from zero at P,Q and R • To a max value at b/4 and 3b/4 • Therefore shear stress may be expressed in the sinusoidal variation in energy through out the lattice
• Slip is assumed to occur by translation of one plane of atoms over another • Reason able estimate of shear stress required can be done by such a movement in perfect crystal • Therefore consider two planes of atoms subjected to a homogeneous shear stress
• Shear stress is assumed to act in a slip plane along the slip direction • The distance between the atoms in the slip direction is “b” • The spacing between adjacent lattice is “a” • The shear stress cause a displacement “x” in the slip direction between the pair of adjacent lattice planes
• The shear stress is zero when • Two planes are in coincidence, initial stage • When the two planes have move one identity distance • Atoms of the top plane are midway between those of the bottom plane
• Shearing stress is a periodic function of the displacement from zero at P,Q and R • To a max value at b/4 and 3b/4 • Therefore shear stress may be expressed in the sinusoidal variation in energy through out the lattice
